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**Hint:**Acceleration due to gravity on the earth’s surface is directly proportional to the mass of the earth and inversely proportional to the radius of the earth. Any changes in the radius of the earth will affect the volume of the earth as well and the change must be counterbalanced by an equivalent change in the mass of the earth.

Formula used: \[g=\dfrac{GM}{{{R}^{2}}}\]

**Complete step by step solution:**

The acceleration due to gravity on the earth’s surface is given as \[g=\dfrac{GM}{{{R}^{2}}}\] where \[G\] is universal gravitational constant, \[M\] is the mass of earth and \[R\] is the radius of the earth’s surface.

Now, the mass of any solid body can be expressed as a product of its density and volume, that is, \[M=\rho V\]

Substituting this value back in our original equation of acceleration due to gravity, we get \[g=\dfrac{G\rho V}{{{R}^{2}}}\]

Earth’s shape being roughly spherical, its volume can be given by the formula \[V=\dfrac{4}{3}\pi {{R}^{3}}\]

Substituting the expression for volume in our previous equation, we get

\[\begin{align}

& g=\dfrac{G\rho \dfrac{4}{3}\pi {{R}^{3}}}{{{R}^{2}}} \\

& \Rightarrow g=\dfrac{4}{3}\pi G\rho R \\

\end{align}\]

Hence, we can now conclude that the value of acceleration due to gravity is directly proportional to the product of the density of the body and the radius of the body, since all other terms in the expression are constants, that is,

\[g\propto \rho R\]

Hence, we can clearly say that if the radius of the earth is being increased by a factor of \[5\], density must experience a decrease by \[\dfrac{1}{5}\] to prevent any change in the value of acceleration due to gravity.

**Note:**If we see the expression for g, at first sight, we might just think that it varies only with the square of the radius and we might not think about the density variation since density does not come into the expression, mass does. Hence it is always a good practice to first express the quantity In terms of the physical quantities whose variation is known to us.

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